Excess electrons on an oil drop, $n=12$

Electric field intensity, $E =2.55 \times 10^{4} \,N \,C ^{-1}$

Density of oil, $\rho=1.26 \,gm / cm ^{3}=1.26 \times 10^{3} \,kg / m ^{3}$

Acceleration due to gravity, $g=9.81 \,m\, s ^{-2}$

Charge on an electron, $e=1.6 \times 10^{-19} \,C$

Radius of the oil drop $= r$

Force $(F)$ due to electric field $E$ is equal to the weight of the oil drop $(W)$

$F = W \Rightarrow E q=m g \Rightarrow$ $Ene$ $=\frac{4}{3} \pi r^{3} \times \rho \times g$

Where, $q =$ Net charge on the oil drop $= ne$ $m =$ Mass of the oil drop $=$ Volume of the drop $\times$ Density of oil $=\frac{4}{3} \pi r^{3} \times \rho$

$r=\left[\frac{3 E n e}{4 \pi \rho g}\right]^{\frac{1}{3}}=\left[\frac{3 \times 2.55 \times 10^{4} \times 12 \times 1.6 \times 10^{-19}}{4 \times 3.14 \times 1.26 \times 10^{3} \times 9.81}\right]^{\frac{1}{3}}$

$=\left[946.09 \times 10^{-21}\right]^{\frac{1}{3}}=9.82 \times 10^{-7} \,mm$

Therefore, the radius of the oil drop is $9.82 \times 10^{-4} \;mm$.